chap003 - .EICCtronic Circuit Analysis and Design....

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Unformatted text preview: .EICCtronic Circuit Analysis and Design. 2’"1 edition Solutions Manual Chapter 3 Exercise Solutions 53.1 flglja 0900 Fora=0.980.fi=1—_'bm=49 Fara=0.995.fi=1;-:':%=i99 49<5<199 E32 a=fi€3=g=m HES-=st 16 E33 7 IE=U+J3JIR So(l+fil=f,—:=E%=SL2S=> fi=80.3 a=lifi=g%=>a=0.98?7 Ic = Na = [30.3)(9.00 0A) => rc = 0.7T1mA E14 a 0.990 5- hymns-99 I I5 -— “5° => [5 .—. 21.50 M 9 = (1+5) " 100 I: = arr; = {0.590)(2.i50) =:~ lg = 2.13 mA 53.5 11-329 rn‘Ic— Ic Ic=0.l mha-ra =1.5 MR 15: 1.0 mAaro=150 kn Ic=10mA=1rra=15kQ 1 Ic—lon l+fi ) => Io = 0.9363 mA A! ch =10, [c = [0.9868)(l + a I —- 1.12 link i 10 0| b ‘4 150 1': =1 =;u(1+ _—_-, Io =0.9934 TBA A! Vcs :10. {C =(0.9934)(1+ ___, 150 1’: =1.06 m.-'\ 53.7 . EV a 200 _ Bbczo = —:‘/%—u = W = 40.0 volts E33 BV BVCEO = C311 :13 va =(U100)(30)=139v 53.9 a. V.=C|.'2<V33[Oll}=’ffi=!c=0‘VO=SV P = D b. V.- = 3.6 Transismr is driven into saturation. 3.5—0‘ - Ia— fl-vfiqlal =4 3111A 5- ch[3at} 5—0.2 _ _ :9. I =10.9 A Ir: RC OM 5—11}... Now that I—C = “,3 = 2.41 < 5 which shows 15 4.33 [hat «in: transistor is Ludccd in mmran'nn. P = [cvcg 4- Java; = (10.9)“).2) + (4.53MOJ) = 2.18 + 3.17 P = 5.35 In“: 153.10 For Vac = 0 =: Va = 0.? V Thcnlc=5gl57=$1c=9JTmA _[c_9.TT_ _ andia— a _ so _o.19amA v; = I503 + vuum) = (0.195}(0.54) + 0.7 = V: = 0.525 V Powc-r = Ich-z + IEVBE = [9.77)(0.7) + L0.195)(0.7) Em“ = fi mW Electronic Circuit Analysis and Design, 2"‘1 edition solutions Manual E3.“ For v; =4 V and {Ca =1.5 mA RC2 => Rr=4kn I _ -V35(°n] -[—10¥ E '- RE 101‘ = —-- = .'1' . 15 (‘00 IC 1.) a m-\ RE:£‘—.°—;'—°-=Rg=s.mm 1.313 —“— E3.l2 lD-Vc _10 —s.34 _ Ic- — RC — 4 => 1c,- -— 0.910 m.—‘\ I _ —V55(0n)-[-1m_l0 -0.7 2, E " R5 ‘ 10 1'} =0.930 [:13 1c 0.915 Ic—alg=>ar—.-I—E-—olgm =3-Cr—D. 8.1-9 I5 = I; —[5- =0.930 — 0.915 5' = mA I: 0.915 ‘3’ [a ‘0015 93““ Va; = Vc -- V; = 5.3-1- (—0.70) => VEE =7.04 V 53.13 _ 10—V55[on) __ 10 —D.T Is- =>IE=L16mA RE 8 15 1.15 = = —-—-— = . 'r' 15 (1+3) 51 #13 22 11A. :3 so _ [c — 1 +513 -— fi(1.15)=fl- IE — Lli [TIA Vc = IcRc — 10 = (1.14)(4} -- 10 = -5.44 Vgc = 0.7 -— (-5.44) =9- Vfig = 5.14 V 53.14 IE _ V35 - Vsahn) _ RE ‘I -G.T =>R5= 1° :R5=3.3kfl I: = of; = (0.9920)(1.DJ =5 1c = 0.992 mA I; = I; — IC = 1.0 — 09920 = Ia = 0.0030 [119. Vac = -Vsc -—- 1cm: — Vcc = (0.9921(1} -— 5 =’ Vic = 4.01 V E3J5 v.93 = IBRB + Vagton} + [53.5 a fgfly + V3,;(onj4- (1 +H'II5R5 I, = fl = i Ra +(1+ 31R; 10 + [76)(1) =3+ In = 15.1.HA Ic- = 315 =(TS}(15.1 HA) ==> [g = 1.13 mA I: =l1+ SUB =(T5}[15.1u.»\)=:» M Vcs = Vcc + V93 - IcRr: - IERE = 8+ '3 - (1.13)(2.5'| —— [1.153(1) Veg = 6.03 V E116 VICE =15 =v V5 = 2.5 V = 15R; V23 = 1335 + Vaswnl + V5 Ia _ V33 - Vadon) — V5 __ 5 — 0.1 — 2.5 R3 "' 10 I3 = 0.15 mA => I; = (101)(0.18) => I; =15.18 mA 2.5 S R = = , = o 5 ls‘lafiflg 0138 m 1389 E317 V55 = 1535 + V£B(0n} + [3R3 2. 13:12 mA=~I5~= £20.0431mA _ I3 .. fl _ Ic —— (1+fi)ls— (51)(2.2) :- Ic -2.16 mA v3... = [2.21m + 0.7 + (0.0431)(50) ==> V53 = 5.05 V Vgc = 5 - IERE = 5 - {2.2](l) =- Vgc = 2.3 V Ella [1) E = IERB + Vas(on]+ I53; (2} 5 = IcRc +Vc5(sa.1) +IER5 11-: = 15 + Tc (1) 5 =1013 + 0.7 + (15 + 1cm) (2) 5 = 4!: + 0.2 + (Ia + Ic)[1) (1} [5.3 = I: +11I3]x s —- 26.5 = SIC + 5513 (3)4.8=51c+I5 4.8=5Ic+ [a 21.7 = 541-3 :5 [g = 0.402 mA From [1). IC 2: 5.3 — 1113 a I; .—. 0.380 mm £3.19 V55- : Vgc-{Sln = 0.2 V H —B.2 - ("5] _ 5 — 0.2 Ic "T 10 =? I; =U.43 mA 0.48 Ia=%‘=—, —w_=_g : v, = —(0.24}{20) — 0.1 => vE = —s.5 v lest Your Understanding Chapter 3: Exercise Solutions E320 E322 I, V; = —4.5 V = V55 < Magma) => Transis‘mr is 3' L3 = b} =_ U‘ h" :13: = {m =1“ = I” = D cutofi'. IE=IQ=IE=CL V£E=10V ‘1' =3? . b. V: = —3.5 V Transismr is active. '3‘ h = a x" Ll! = 0' IE” = In 2 0 5 —0.T _ V; = IBRB + V35[on) + IERE — 5 [m _ 0‘95 :13] .- 4”; mA ‘ _ '3 5 — 3.5 = IBHD} + 0.7 + (75}1301) la: a 0 and. 21c] 2h; = 3 mA 5-15—03: . '. = .__..___.._._ = I - 1» =02 \ 1; “WWW! 00025.: mA 0 #IE=Z_53#A c, V1=Vg=5‘~'-. IE§=JI32=4.5311L‘\ I: = fiIg = {75)(255 3A} => 1.: 7: 0.191mA In = 8 mA. 161 =16: = 4 IRA! VB = 0-3 V I; ={1+ 5)]; = (76)I:1.55 MA) E3 23 = IE = 0.194 mA I . VCE=10-IcRc-I5RE v0-5—rcRc=S—fisRc and =10 —(o.191}{2}-(o.194)(4) i = V5,, +Av, —V,E(on) Va; = 5.34 v " R3 Then c. V. = +3.5 V Transistor is in saturation. Av _ _flRcAvl a _..__........ R (1} 3.5:13R3+V55(on)+IER5—5 or " [2) 5 = IcRc + VCEBM} + J'sz — 5 Ave _~ (3) IE =13+Ic Av, Ra Let fi=100, Rc=5kfl, Rfl=100kfl (1‘13.5+5—D.7=1DI&+4(I5+1c) Then [2) 5 + s — 0.2 = 2n: + ms + :c) Ava ‘ -(100)(s) _ S (1) 7.3 = 1415 4.41,,- Av, ' 100 ' [2} 9.3 = 415 + EIc Want Q-paintto be 3 x [1) =- 214 = 4213 + 121C "0(Q-Pl)=25=5_(100)‘{39(5) Then me =- 19.5: SIE+12J§ V -07 =0.005mA I =0005=L 3.3 = 3“? .LL—‘ “9 100 =I§ =D.li'2 mA (I); V =12 7.8 =14(U.112}+ fife = 1.553 + 416 "L a I = L56 mill Also 1Cg =13!” =(1oo)(o.oos) [c "_ 0r E:l$.9<6=¢1nsaruratiun. ICQ=05mA IE=I3+Ic=s-f§=1.57mA E32 Vcs = 1"c-,ns:(5a.r.)= 0.2 - 4 a. For Vcsq = 2.5 V 5 [ca = 5 -22-5 =:- IcQ =1.25 mA 5-15-02 I sat ==———-= = 9 c( ) R 15::- R 0.22014) hm = I? = ___1l-D-05 z, 15;, = 12.5 pA 1c 15 5—0.3 ' - _ ~ Itz—02—0=0.7SMA= R, Thgnflszznlozgfiflfi=344kn or . RB :55“) b. 159 15 indcpcndcm of 13. For¥c5q=1V, Ic=3_1=2mA Electronic Qircuit Analxgis and Design, 2"d edition Solutions Manual 1c 2 - = __ = 1 =150 _ 1-0-r _ '3 In .0125 a 3 130 —- = 139 —17.6 #3 . . 5 - 4 _ . For ‘11ch = 4 V, 1c = ., = 0-0 mA Icq = 131'“ = (35)(17.6 11A] L: 0.5 :3 1 =1.32 11114 := —— = 3 = 0 3.9—— ‘5 13 0.0125 =" “ 120 = (1 +1311“ = (?6](l7.6 11A) => IEO =1.34 111A E325 Vase = 5— (1.321(1) — (1.3410121 5040<d<160 ._0-- =>V55Q23AIV ° 800' = In = 0.005355 mA ———*"—" 13 = 75 =5 {an = (751100053751: 0.403 111.1 E317 :3 =150 =1- Icq =1150)(0.0053751= 0.606 11131 Largest Icq = 5111311051 Vcsq 5 —- 1 = $.96 = 0.506 m 9-: ‘K 4 ° ' = 2.45 m 0.405 I39: ‘21:!“ 3:150:35: S=TS=~RC= For a nominal Ic = 111.604 mA and 17.35.; = 2.5 5 — 2.5 RC ‘ 0.604 For Ice = 0.403. Rfagx chq = 5 — (0.403)(4.11) = 3.33 V For Icq = 0.606. Vega = 5 - (0.306)[1.14] = 1.66 v 50 fox R: = 4.14. 1.66 V 5 Vcaq 5 3.33 V = 4.14 = R: 1' R; + R: =11.25 kn. 5 =15!) Ic z Ia. Vega = 2.5.V So Icq :..- 15.; = = 2.081mA E326 [so = 16?" = fig =13.9 1.111. n. Rm = Rlflfiz = 9||2.25 =5 Rm =16 1:11 I _ V” _VE£[W) V R: “5 5" Rn: +11 + 51R: 3'” = (m) V” = [9-1- 2.25)“) : -u'——V = 1‘0 v (*RITEZ) 'cc - V5503“) 0.0139: 1 2 h. I“ = v” — Vezfan) z 1— 11.7 m +11 +5135 Rm +11 + 15mg 1.6 + (151110.21 R; )6) * M =# Isa = 9.375 111A _—_ Riggs Ice = 31's.; = (150149.375 11A) ([125] +(1511(0»2) =>Ico = 1.11 mA I30 ={1 +13)I$Q 515:: = 1.42111}. Vcsq =5-Ichc-IEQRE R2 = 11.25 —- R1,. 50 0.0139[R;(11.25 - R;]+ (15111021111251) = 5 —(1.»111(1) 41.421012) V = 3.31 V , .. -——'9————” =a(11.25—R1) 40.111125) .3 pm- ,3 = 75 0.15631 — 0.0119113 + 1.72 = 56.25 — 511, — 7.575 0.01393? —- 5.15611. + 43.66 = 0 =53; - (0.1)(11251 5.156 i (5.156)’ — 4(0.01391{43.561 _ 2(00139) => 34: 8.67 kn wd R1 = 2.55 m R-l Test Your Understanding Chapter 3: Exercise Solutions E23 .28 I dc equivalent circuj: :3 =150. Rn: = RIHRZ. Vary = (—13—)(1m—5 RI+Rz 5—0 1:4: 1 =5m.'-\ fax; 5 I =—=—-—-_- ‘u so a 150 0333mA VTH-t-S-UJ 159- ‘ Err-1+0 +5)Rz Sc: RrH = (0.1|(1 +.5)Rs I __ V111; +4.3 3° ' (1.1)(1 +51Rs R: _ (R: + 32):”) — 3+Jt3 : [10333 = (1.1)(151n02) (0.0333)(1.1)(151)(0.21=(%D(1n)— 0.7 R; _....._—. = 0_ 30 (R1 + R2) 1806 Rm — T— : {0.1p{151}(0.2} = 3.02 m Then R.{0.1006) = 3.02 =- 121 = 15.7 m a, = {0.1805}(16.T + R1} = 0.31903; = 3.02 :9 R; = 3.65 M! 53.29 :3 =120. Vcsq = 5- V . R: RTE R1llfie. TH (121+ R2){ID] 5 In; == Isa 10 - Vcsq 3-: + R: “3-5 1:4 — =>- Icq = 3.33 mA 50 Ice = Isa: LEE-=— %:—:-=? {30 =0-0273 "‘A v” +5 -0.7 .r =——-——-— Ba Rn: +(1 +5135 .56! RTH = (0.1)“. +3313; R: . — —D.-' (R1+R1){10) 5+5 [1.13(121)(0.3) 1'90 :— D.0278 = [0.027a)(1.11(121)(0.3} =( R” )(101— 0.7 Ri+R2 (-—£’—)—0131 RI+RI "- . RIR: _ _— _- . = 0.! l . R73 R1 R2 lefl 131) ( 21)“) 3] =:- R; = '20.} in R: =(0.131)(20-1+ Ra] =:» 0.01%; = 3.03 m fi' R2 = 4.44: kg £3.30 5 “ {100 = = — = .99 A a- fag (1+E)IEQ \10])(1} U m 1 mA 1 = —--— = — .90 . [5" 1+3 101:"9 “A '5 = --I3qR3 = —[0.UU§9)(50} ==> Va = —D.495 V . _ . Icq _ I 0.99 x10"' 03,; _ vr1n( IE ) _ [0.0263111 («———-s x 10m ) =. V55 = 0.530 V v5 = V3 _ V55 = 43.495 _ 0.530 =9- V; = “1.13 V v.5 = 10 - (0.993(3) = 5.05 \' Vcaq = W: — V; = 5.05 — (-1.13; 201-5}; = 0115 v b. [sq = 1111A. 1-3 = 51—] =U.0195 mA v5 = 400195150] = —0.93 v = V5 13.; = (553%)(1) = 0.95 mA 0.90 x 10-3 3 x [0"“ V; = -D.95 - 0.629 = -1.61 vs =10 - (0.95)(5) = 5.1V Vcsq = 5.1 — (—1.61) =- chq = 5.71 v V55 = (0.026|1n( ) = 0.629 \' Electronic Circuit Ana} sis and Desi 2"“ edition Solutions anual E331 E133 ._ .33.. _ IS. .. ’_° _ {a _1+ _ 121,113 .. (121)120) _ 11310155} 1/E = IQ10.105) +0 1 20 [co =(1ffl)rsq = In -—- (0.992110 VC 2 ICQRC — 5 = {0.992)IQ(4) — 5 = 3.97Iq - 5 VECQ 2 VS — VC = [19(0105) + 0.7] — {3.9719 — 5] =‘ 4.80512; + 5.7 — 3.80511; + 5.7 = 3 . =0 IQ = 0.710 mA I _ VrH - VBE(°“} _ 4 ___u'7 E" ‘ 12m +(1+31R21 ‘ 33.3 +(101'1121 131:14 0101. 1C; =1.40 1119.. M. E332 v3: = 4 —(0.011)(33.3) .51: 45V :9 E] = 3.53 V. V5; = 2.83 \" IR+IB? =Ic1 12 —- Va 12 -(Vc1+0.7] =’ 5 + (1011(2) =1'“ 12 (12—07) _h Va. ‘5"+ 11011121 “"w‘ 5 +(101)(2) 2.4 + 0.0559 — 1.40 = meq + 0.00495) VEI = Va; = 5.15 V. VE; = 5.35 ,, _ - [5: = :5 [52 = 3.105 mA. Ic; = 3.04 mA. 132 = 30.4 11.4 Va; = (3.041115) :3 11-, = 4.55 V R7}; = SOHIOCI = 33.3 kn . 50' - VTH —- (m)(10) —- a — —-1.67 V - - 1.67 - 0.7 9.63 I = -a——-—---—- = :— = .3 . 9' 33.3+11011(21 235 1‘ “" Icl = 1.12 mA. 1'51 = 1.13 1112‘. V31 = IE]. H5: — 5 ={l.13}(2} — 5 = —2_74 V ch = 3.25 V =- Va; = 0.51 V =- Vgg = 0.51+O.T =1.21 V - — .21 IE, :7 =1.90 m}; =3 1‘3, = 13.3 11A 1.3, = 1.30 111.”. Im = In — 1m =1.12 - 0.01311 = 1.10 mA 5 —0.51 RC1 —- =9 RE; —l.DB kn VEc: = 2.5 =- Vc: = V17: - VEC: = 1.21 - 2.5 = —1.29 -1.29- (—5) Re: = L83 =5- Rg: 21.9? 111'! Electronic Circuit Analysis and Designl 2"“ edition .___________501Uti0“5 Manual Chapter 3 Problem Solutions 3.l 3.2 (a) Patti—2110: G=L=m=099099 1+.5 111 ' Forfi=180: 180 G — m — 0.99915 0.99099 < a < 0.99448 (b)!C =fi1,=ll0(50p4):> 1c =550mA or I: =180(50p,4)=>. Ic = 9.00m SD 5.50 s [c s 9.0 mA 3.3 5=ICR+VCE - V35=Ic(2]+1-0.7 =:~ Ic = 2.35 mA Ic 2.35 15— a —m=>fg=2.39m2°u 3.4 Same Figure as Problem 3.3 5—1.3 vc =—0.7+1=1.3V', ic=—1-——: it. =l‘85mA ' LBS . For fiF=120, :B=—'£—=—-=> :,=15_4m ,5; 120 {E {1}” )jc =[12_1)(1.35):> is =1.865mA p 3.5 3.6 “a I} h... ‘h‘ E 3 II 5. ‘6’: it *3 fi U “In. II to :4 H1 3 :‘C = [fl}21_7) = ic. = 27.4 In»! ' 27.7 is: ’5 =——=~ i5=0_304mA up 91 —— 3.7 DBVECB 12 is = Sle'flp’f :3 05:10-3 = Sleufisofunza So that [,1 = 6.94:10'” A Device 2: 12.2:10" = [522“sz Or 1‘,1 = Lame-'3 A I 1.69:10'“ . Ratio afareas = i: —--—::> Rang = 24.4 15. 6.94x10'” 3.3 r V (a) r,=—-"—=2—S—q:> rfl =250kfl Is I F 250 r=—"=—= =2‘SOMQ (b) ' 10 0.1 r“ 3.9 Bcha 50 V = = — 3 “9° an? m BVC = 12.9 V Electrgnic Circuit Analysis and Desigg, 2“" edition Solutions Manual 3.10 3.“ 3.12 BV BVcso= £0 -251 a -fl- 56... W»fi_ 56 _3.93 g=eo.s Bv BVch= $38” 3v¢m=wvcmfl=mw€fi I BVQE§=154V (a) 15:121‘00'7 => IE=1.13mA 75 I = — 1.13 =1.12mA c [76 3 I»:=5 = 24 '- (1.13)[10) —(1.12}Rc = a so that 1'?c = 5.98 kg 1 I =—=0.0132mA (b) , ,6 V, = —I,R, = —(o.0132)(so) => V, = 4.653 V Ic = [E 1) = 0.937 mA 76 5—2 =3“: 2 RC 0.937 L 1041‘“ c_ ;3=w:15=111w‘ 10 +[76)[10} I: = 0.905 mA V; = 0.7 +{o.0121)(10) — 2 V; = -l.lfl V V;- : IcRc — B = [O.906){3) — B = V: = -5.23 V Vic = V; — Vc = —1.15 -— (—5.25) n» Vgg = 4.1V :1. 5 =(1+ 3)!3(10} + 13(20) + 0.7 + (1 + BNBR} 5 = {BUN} +20 + 152] + 07 =- 13 = 4‘61pA' Ic- = 313 = US$1.61; :- Ic = 0.346 mA V; = s — (1 + mime: = .5 — [TE)(0.00461}(10) = V; 21.50 3.13 (a) Figure P3.12(c) 3=(76}1,(10)+o.7+1,(I0)-—2 I = 10—0.? -' 10+(76)(10) Then ' 1C=(75)1,=a 1C=0306mA and 15:0.9I8mA = 0.0 l 208 mA Vac = 8' [5R5 " {CRC "('8) vac = 16—(0.918)[10)—(O.906){RC) vac = 6.32-—(0.906)RC RC: 31:91:593 = 2.355 RC glism Then 3.9? _<_ VEC S 4.24 V (b) Figure P3.12(d) 5=(l+fi)I5RC+I,(20)+OJ +(l+fi)15(2) 5 = (76)],RC + 1,00} + 0.7 + (76)IB(‘2} Now RC =10k£1i5%= 955 RC 5:051:31. Then 0.00443 5 I, ~_: 0.00481mA And VC =5—(1+fl)1,RC So that L46 5 VC 1:153 v 3.14 V -v 2.5-0.7 R =M= =9 R =120kfl " I, 0.015 ‘—._-_—. Im=(70}(15,wl)=>1.05m v —v _ Rc=fl=flm RC=238kfl [ca 1.05 —— 3.15 _ _ -Va __ -(-1} a, V5——15R3=>IB— RB — 500 Is=2flp5n ’Ez—l-CI.T=—ITV EVE—(n3)_—1.T+3_ 7 15.. RE .. I” -02 08mA 1E _ _ 0.2703 _ , _ _ E _. (1 +13) _ 0.002 —139.4 = ,3 — 134.4 .. 5 _ a—1+‘B=:'g—fl.9926 1c =Bfa =9- Ic =0.259 mA Va: =3- VE=3—(-1.T}=>VCE :43? V Chagth 3: Problem Solutions 3J6 b. In: i¥=575=fljmls 4 = 0.7 + 1335 + Us + chflc — 5 In + 1c = h; 4 = 0.7 + {3(100) +{0.5}(B} — 5 I 0.5 I; =o.o43=:. —E= (HI!) = ---— =11‘63 [5' 0.043 13:10.63, a: 13.6 =>g=ll9140 I. Va=0=>CutDfi=9fE=0,V¢-=EV -.7 b. V5=IV,I£=l ° :5 Ig=0.3mfi. fr: 2 I2 = Vc =5—{0.3](10)=> V}: = 3V c. V3 = 2 V. Assume active-mod: 2-10“=I£=1.3InA=Ic Vc = G — (1.3)(10) = —7 V1. Transistor in Man 2 -- 0.7 Is: f5: = = I; = 1.3 IDA Vs =1.3 V. ch(sa.t] = U32 V Vc = V; + Vcsisu] = J..3 + 0.2 =¢V£=1.5v b. V53=1V 1—0.7 I:— $0 $611!. fc=fl[5=(75)(5)%fc=fi.45mA S-Vo_ E s 'I°+10 1—0.45=Vo(%+ :Vg=1.33V c. Tmsislor in “Malina Va "—1 “cabin = 0.2 V 3.13 (a)/3p=100 . 100 (1) IQ=OimA 1c {Tb—l o.|)=o.0990 m _ V0=5—(O.099)(5):> V0=4505V 100 u = I = -— 0.5 =0.495MA (11) IE, 05mA c [101 ) I Va = 5 —(o_495)(5) : Va = 2525 V (iii) IQ = 2 mA Transistor is in saturation V0 = —VBE(sa!) + VCE(5ar] = —018 + 02 z} Va = —0.6 V (b) m = 150 . ' 150 (s) 1‘? = 011m 1}: = [ET 0.1) = 0.09934 m4 V0 = 5—(0.09934)(5) :5 Va = 4503 V Va change = Marlow/3 = -0.044% 4.503 .. 150 (u) 19 = 0.5 mA I: = a 0.5) = 0.4957 mA VD = 5—(0.4967)(S) 2) V0 = 2.517 V % change = Marlow. = —0 32% 2525 (iii) IQ = 2 mA Transistor in saturation Va = —0.6 V No change 3.19 I£=VEIOJ a 50 5—1/6 :2 = — V — = k ([+3)IE (51)“? 0'7) m IMVc=VE so . _6-V5 (51)”5 0'7)“ to 9.300}, - 0.7} = e — v5 10.30% = 5 +{n_7){9.501=. Va =1.19 v __ 1.19 — 0.7 13‘— —"'T""'-=>I5=0.49mfl. 3.20 S — 0.5 Vm=OjV=> VO=D.5V, 1c: =0.90mA 101 I = —— 0.90 = 9 [mo ):> IQ 0.909mA Electronic Circuit Analysig and Design, 2'“ edition Solutions Manual 3.21 10—V5_10-2 _ IE_ 10 _ m =>IE_G.80 mA '3=Vg—0.T=2—O.T=1.3V V3 13 f3=R—s=fi=>lg=0.026mf\ [c 0.774 = .....= __ = "7 fl 3 0.026 :5 29' 0 _ 29.7? -- .£I'—[_+a 307T=>a—096m V5: 2 V5 - vc = V5 - {IGRC — = 2 — [(0.774)[10} — 10] V5; = 4.26 V Load line developed assuming the Vfl voltage can change and the R, resistor is removed. I 11114 ._ -, 1;; 0.77% ‘9 39° " fl; LL LG Vé‘: 2.6 3.22- so = — 1 = 0.90 mA 1,: [51} 1 Va = IcRc - 9 = (o.93)(4.7) — 9 or c = 4.39 V I, = lf=00195 mA Vit = 1.12, + V3001) = [0.0196)(50) + 0.7 or , = 1.68 V 3.23 so 05 = —— =0.49mA.1 =—=o.0093 A [c [51 05} , 51 :11 Va. = 1,11, +VE,(0n) = (0.0098)(50) +0.? or V‘. = 1.19 V Vc = c126 - 9 = (0.49)(4.7) — 9 = 4.701? Then V” = V, - V: =1.19—(—s.7) = 7.39 V P,2 = [CV55 +131!” = [O.49)(7.89) + (0.0093)(0.7) or Pa = 337 mW Power Supplied = P, = 19(9 4,) = (0.5)(9 — 1.19) Or P; = 3.91mi? 3.24 For IQ =0, then PD =0 For 1.2 = 05 mA , [C = [3}- 05) = 0.49 mA 51 I, = = 0.0093 mA, V, = 0.490V, V5 = 1.19 V VG = (0.49)(4.7) — 9 = 45.70 V 2:. Vfit = 7.39 V P 5 1?ch = (0.49)(7.39) :> P = 3.37 mW For IQ = 1.0 mA, Using the same caicuiations as above, we find P = 5.95 2213’ For 12:15:71.4, P2626mW For19=2mA, P=4.80mW For 12:2.SmA, P=157mW For IQ = 3 111.4, Transistor is in saturation. 0.7 +I,(50) = 0.2 + 1601.7) — 9 IE =IQ=15+IC=II=3—IC Then, 0.7 +(3 — 1C)(50) = 0.2 + {C(17) — 9 Which yields 16 = 2.915 171/! and I, = 0.034 01.4 P = [EVE + 1’ch = (0.034)(0.7) +(2.916)(0.2) or P = 0.642 mW 3.25 IE1=I32=§=>IEI=IEQ=Dij I51 = Ic? z 0.5 mA Vc1 = Va: = 5 - [0.5){4) # VG: = Vgg = 3 V 3.26 a. [so = ——Hc -VB£(°n RB I59 = --—- = -a = 0.0333 01A 24—0.? Ra - 00333 = R3 — 599 kn Vcc -Vc£q _ 24—12 RC 1:» RC — 2 m-Rczfikfl log = _ Vac — V35(on) __ 24 - 0.7 1:. I39 —- RB — 699 = 0.0333 ILA (Unchanged) 1m 2 1015.: = [100)(00133) =? I'm; = 3.33 mA _. Vceq = Var: — Icoflc = 24 - (3-3)“) = Vega = 4.02 v M. gghaptcr 3: Problem Solutions (:3) was = Vcc - ICRC z 24 — [C(e) 3.27 IE = VEE —V53(on} = 9 —-D T as 4 => 1: = 2.075 M Ic = 1115 = [0.9920)(2.075) =19 IQ = 2.05 mA Vac +IcRc = Vcc '3: = 9 - [ZIUSNZJJ =5 VEE = 4.47 V 3.23 _ — Va; 12 — 5 Ice — T = = 2.73 mA [cg 2.73 [sex T: 3-0—3 Ifiq =U.091mA 0.7— (-12) I = = R2 mu 0.127 mA In; = Im + {sq = 0.127 + 0.091: 0.315 1.11.“. V. = In: R. + 0.? =(u.21s)(15)+ 0.7 = V; = 3.97 V 3.2.9 For Veg = 1.5 5 —‘l.5 Icq = I = 0.5 mA 0.5 [sq _. Eg— = 0.02 IRA 0." — — 133 = —‘Tgsl = 0.057 mA 1m = {R1 + 19:: = 0-057 + 0.02 = 0.077 mA V. = In; R. + Vas(on) = (0.01?){15) + 0.1 =1.35V FWch=1.0 1cq=a:1=-imA 4 I =-——=0.6 an 25 1 nut [33 = 0.057 M Im = In, +I5q = 0.057 + 0.16 = 0.217 mA V. = {0.2171(15) + 0.7 =- 3.96 v So 1.56 < V1 < 3.96 V 35:. q. ‘1' (544%.; or: Q'fA‘. VJLUEJ; a; -— o 1 «ar- y 3.30 R“, = R1IIR; = 33||m = 7.67 m R: . ID V” ‘ (R. + 3.)“ ‘(1o+33){15} = 4.19 \r' VT}: — V35(Dn) _ 4.19 - 0.7 Rm +[1+ m3: _ 7.5? +(51)(1) fag = 0.0595 1111’; Leg = 160 = ISIBQ ‘4' Ice = 2.97 mA [sq = 3.03 mA Vcsq = 'cc — Ichc - 1'squ = 1a . [2.97)(2.2) — (3.os)(1; :Zsfliflfl 3.31 ImzllmA, ch=9v, Rm=som Also I,=-§-g~=0.015mA V”, = {mm +V,E[on) +[l +fi]IaQRE V”! = [Eff—R1}Vcc) = Rm 'Vcc = "i"{so)(13) Then éfioxm) = (omsxso) + 0.7 +(31)(o.015)(1) - 338R. - or R, 4538“). Then m-SO: R, =ss.7kn 15:, {$3 1.2) = 12151111! 18=Ich +Vm+meE 18 =(12)Rc +9+(1215)(1}=> RC = 6.49 m Elcctronic Circuit Analygis and Design, 221"d edition Solutions Manual 3.32 v” = (&){24} = 9.5 v 39.9+60.9 R,,,=R.|;R,=20fi15=057m 95 D_ __. —' ‘ " = 0.00553 A V = R? V )=[ 15 )(lolmtzw I“ 24.1+{126}(10.5} m m R. +R: cc 15+20 [cg = 0.317 mA, 13.; = 0.823 mA I = _. r v =15.w Vac ___ I.ng +Vfl(on] + £9 +Vm VCEQ 24 . (0323)(10 a} 3 can 50 0.817 5 [cg 51.02 mA and 10: rm(1)+0.7+fm[al'fi]+4,29 14.2 g ch g 10.1 v. The“ 3.34 -O.7— . . R =R R =258=6.06 in I =10 429:5 l:> I =4‘62m‘4 3 TH 1|] 1 I] “9 +331 1.035 ia____ V _ a? )V _ s \[m 101 m“ R1+R2 m" s+25 I . . fill-Rm +V;W=[1§‘i 357)+4.29 =5.52\ 1+}? 101 I _ a”, “magma; = 5.52 — 0.7 0r 3‘3 ‘ Rm +(1+ am; 5.05 +{76)(1] ml: 13¢ = 0.0524 mA. Icq = 4.50 mA 3.33 159 = 4.74 a. Rn; .= mum = 531142 = 24.35 km Vcsa = Vcc - IcaRc - Isa Re : '2 — . 3 - 4.74 '1 V _ R: \V y 42 24 i .4 man: i JH 7"" 3:4:sz CC“ 42+sa ( ) ___"CEQ=°-12V = 10.1 v 5.52 — 0.7 b_ r = —-———— I = [1.0326 mA I _ Fm — Vsflon) __ 10.1- 0.? 5" 5.06 + (151x11 : i"— B" “ Rn: + (1 + 5mg "‘ 24.35 + (126)(10) I“, = 4.39 talk by; = 0.00732 mA [sq = L92 1:0 = M80 ={125M0-00732J chq = 24 — (4.39)(a) — (4.92m) => [co = 0.915 [11A chq = 4.41.“. 15.; = 0.922 a chq = 24 — (0.9223(10) 3 35 =>Vc50=14.sv (a) [WEIRQ=0‘4MA a. Let Rc=i=>12c=15m; RE=—3—:>RE=7.5kQ R2 = 4:2 ta + 5% = 44.1 m 0.4 0.4 9 R:=58kfl-5%=55.l kn + 5—.—=1125k{1 R‘ R’ (0.2)(0.4} R5 =10kn—5%=9.5kn Rru = min: = 24.5 m pm =[RiiiRjyoJ = [593,” +V,E(on)+(1+fi)I,QRE «.1 = —.__ = , 112.5— Vm (“'1+55.1)(24) 107v Rm Rd?1 _( Rama, 10.7—0: "R,+R,_ 1l25 LEG = =0.00319 I =0i=oflo4m log = 1.02 mA. I“ =1.03 mA ’9 100 ) . [125—111 R, VCEQ -"'- 24 — (1.03)(9.5] => chq = 14.2 V 9 ___ (_______,_ +0]! R: 112.5 (0.004) 112.5 R L“ m y n 4100(000405) = {2 -— = . a 5 o 39 g k We obtain R5 = w m + 5% = 10's m From this quadratic, wcflnd Rrazfl1llflz=24.lkfl R1=48kQ =2! R1=545m Chapter 3: Problem Solutions 0)) Stande resistor values: -Sct RIr =Rc =15H1 and R, =62kS1 R, =47m Now R". = RallR‘J = 62||47 = 2mm _ R2 _ 47 _ Vm {RI +R1)(Vm]-[4?+62)(9)_ 3.331! Vm = [IQRTH +V2£(O"}+{1+fl)laQRr So I — 338‘“ —000406 A ’9 26.7+(101)(7.s)" m Then 49:03:06)“ pm = V“ =(0.406)(7.5)= 3.05 V 3.36 1- RTE = Kill-R: =12"? = 1.71 = HIE . R2 2 v = — = _ ’ TH (R1+R2)(10) 5 (12+2)(10) a = -—3.ST V = V?” b. V —- _. _ 13‘; = TH Vasfon) ( 5) Rn; + [l +13)R£ _ —3.57 .— 0.7 + s 0.73 - "" =- Iaq = 0.0140 111A Ice = 1.40 mA, Isa -'= 1.41 mA. chq =10 -- Ichc- - IEQRE =10 — (1.401(5) - (1.41)(0.5) ch = 2.30 v d.For R: =2kn+5%=2.l m R: =12 kfl-5%=11.4m R5 = 0-5 m — 5% = 0.475 m Hrs = Rallfia = 1.7? tr! 2.1 V73 -- (2.1 +1I‘)(10) -5 -- -—3.44 V —-3. —0.7 5 . I“ __ . ____........_4" + — 9—55 =0.0173 mA ' 1.77+{101)(0.§753 " 49.7 1.3;; = 1.73 mA. {sq = 1.T5 mA F01ch =5kfl+i%=5.25kfl Vcso =10 — (1.73)(5.25) — (1.1301475) =0 V539 = 0.0863 V (Saturation) F0! R2 =2kn-5%=L5m Rl = 12 kfl+5% =12.6 kn R5 = 0.5 kn + 5% = 0.525 kn an = R; "R: =1.55 kn 1.9 V = —_ fl _ a _ _ V T” (12.s+1.9){1 ) 5 3 69 —-3.69—0.7+5 0.61 I = — = —-- = , 3" 1.65 + {101)(0.525) 54.7 0 0112 I“ [co = 1.12 mA. Isa = 1.13- In}. For R;- =5 kfl-S‘FE =i.75 kn Vcaq =10 - (112101.75) - (1.13)(0.525] =-.~ Vcaq = 4.09 v $01.12 5 leg 5 1.71 mA and 0.0863 (V < . V. _ cgq _ 409 Saturation 3.37 RTH = R1HR2 = 9H1: 0.90 kn Vm =(R1‘ERQ)(-12)=(llfi)(~m = #1.? V IBQRE + Vgafim) + [Hafiz-H 4- V7}; = 0 Mn.) _ 1-2 -' 0-7 Rm +(1 +0135 ‘ 0.90 + {76)[0.X) [30 = 0.9533. Icq = 4.41mA Lac = 4.47 mA fag: Cantuaflmdlinc: Vgcq 26V 1293-: + Vscq + h:ch- -12 = U (4.17)(0.l) +6 + (unm- =12 av R; = 1.26 kn Electronic Circuit Analysis and Design, 2"" edition Solutions Manual 3.33 (a) R”, = (0.1)(1 +133; = (0.:)(101)(05) = 5.05 kfl 1 VT" = K. Rm .0“ = 1,077," +V,£(an)+ (l +fi)7,QRE l I . ,r —.£G_=fi=o,oosm ‘9' 13 100 Then $505710) = (0.003)(5.05)+ 0.7 +(l01)(0.008)(05) OI.’ .R RI =44'Em‘ £41; 44.! + R2 New {£9 = [191)(03) = 0.303 m 100 vac = Ices: Wm + 1593. 10 = (0.8)RC + 5+ (0303)(05) RC = 5.75 k9. = 5.05: 132 = 5.70 7.9 (b) For 75$fiS150 _ R, _ 5.7 _ vm {RI “a: )(VCC) _ (—-—5‘7+M'1](10}- u45v VT" = “QR”, +V,£(0n)+(l +13)I3QRE _ l.l45—D.7 ‘9 7 5.05+(76)(05) Then [CO = (75)(0.0703) = 0.775 m For 13:75. I =0.0103mA 1.145-0.7 For = . I = = fl ’9 5.05+(151)(05) 00552 “A Then [CE = 0.829 m AI .. 95 Change = m = 0529 0375 x100% =- Icq 0.80 96 Change = 6.75% (a) For R5. =15] Rm =(0.1)(10i)(1)= 10.1730 Vm =“Rl—' Rm 'Voc =%‘(10.le0) =- l (0.003)(10.t}+0.7 +(10:)(0.003)(1) which yields RI = 63.6 kg 63.617, 63.6 + R2 12 NOWV = R2 V = = ’" [almjcc] [12+63.6](10) '58” 1587—03 F =75. I = = _ And =10.1==. R, = 12.0139 So IQ = 0.773 m l537 - 0.7 FD =15 II = “6 o '9 10.1+(151)(1) = 0.0055! mil 3.39 3.40 Then Ice = 0.826 M Mm _ 0326 — 0.773 [W ‘70 Change = 6.63% % Change = x500% =3 Vac a Immc + RE)+ch 10 = (0.8)[Rc +R£)+5= Rr + RE =6.25 7.9 Let RE =1kfl Then. for bias stable Rm =(0.l)(121)(l) = 12.1 kfl 0.3 I” = m = 0.0066? 77174 $41ti10) = (0.00667)(i 2.1)+ 0.7 I +(121)(0.00667)(1] 76.2771 30 R, = 76.2 m and —— 76.2 + 772 212.1: R1 = [4.47.0 Then 1’ "Hm-Lo— ” ‘ 76.2+14.4 This is close to the design specification. =0.l 10:77.4 Ice = Lee = cho = Vcc - Icqlfic + RE) 3 = 12 — Icqm + 0.2) 1.5:; = 2.73 31A. 13¢ = 0.0213 mA Veg-q = a v VIN = IEQRTH + Vaalon) + (1 + fiNaQRg -6 . R: ) l? = 12 - 5, R = R R TH (Ri +32 ( ) TH I“ : Bias smblc=> RT}; = (0.1)“ + 3)]{5 = (0.1)[125)(D.2} = 2.52 kn my = (3%?)(RTHX12J-6 i(2.52)(12) — 3 =(0.0213)[2.52} + 0.7 R: + (123)(0.0213}{0.2) — 3 343024) = 0.7549 + 0.5494 R1 23.2122 — ° '1 —- = 2." M- 213+}?! 32 R2 = 2.33 kn _. _ _ ghaptcr 3: Problem Solutions 3.41 80 I. Ice = 1 mA. 15:; = )(1) = 1.01 mA VCEQ =12 - "- =D VEEQ = 9.30 V 15,; = 31—6 = 0.0125 mA R” = +[o.1}{1 +5112; =(o.1)[31')(o.2)=1.62 m R: _ l a _ P'ng (RI+R:){12)—6—R—I(RTH)(i-) E 1 _ R—1(19.44}—- 5 VT” = {EQRTH + VBE(OI§) + (1 + 15)IEO RE "" 6 RLuwm) - s = (0.0125)(1.62) + 0.1 1 +(31)(o.0125)(o.2) — a ill—(19.44) =o.923 1 _ 2!.132 R! - 21.1“), -- 1.52 R; = 1.75 kn b. R; = 22.2 1:9 or R1 = 20.0 H"! R: =1.“ 1:0 or R; = 1.66 kn R5; = 0.21 1:5"! or R; = 0.19 kn RC = 2.1 kn or RC =1.9 kn Rfimu}. Rdmin). R5(m‘m) Rn.- = (1.34}:|(2n.0) = 1.685 m 1.84 V _ _.._._ _ — _. TH (LB‘H_ 20_o)(12} 6 4.99 V 4.99 —o.7 — (—6} _ 0.31 1'.635+(81)(0.19) ‘ 17.05 Icq = 1.45 mA For max. 12.: a Veg = 12 -—- (l.45)(2.1) - (1.47)(D.19) VCE = 3.63 v Isa = = 0.0182 nm R2(nfinJv R1(ma~x)l R“,- = (1.sa)||(22.23 = 1.547 m 1.66 v = _ _ =_- _ T" (1.66+22.2)(12} 5 a-lfia'v’ -5.155— 0.7+6 0.135 r = ———_.__ = __ 3" 1.547 +(31)(o.21) 15.55 = 0.00727 11A For min. RC ==» 159 = 0.53: mA, I; = 0-539 Vcsq = 12 - (o.552)(1.9) — (o.sas)(o.21) chq = 1037 V So 0.532 _<_ Ic 5 1.45 m 3.65 S Vcsq 5 10.77 V 3 .42 3.43 Vm2 Eva —IW(RC +RE) 5: I2—3[Rc+RE)=> RC+R£ 22.3316). but R5 =0.33m and RC =2m Nominal value of A3: 100 Rm ={0.1)(1+ @125 = (o.1)(101)(0.33) = 3.33 L9 [B9 =%0=0.03m.4 l l .. pm =E. Rm .92)- 5: ET(;.33)(12)— 6 Then Vm = 1,91?“ +Vn[on)+(l + fiNBQRE — 6 Ri(3.33)(12)— 6 = (0.03)(3.3 3) + 0.7 + (101)(o.03](o.3 3) — 6 which yields R. = 22.2 kg and R1 = 3.92 kfl Now R 3.92 V _—_ 1 12 -6: ——.__ 12_ ”' [3%ng ) [1924422) 3' 6 Or VT" = 4.20 V For fl = 75 , Vm z IsaRrH +Vx£{on)+ (1 +16)IBQR5 ‘ 6 I = vm +6—0.7 = wane—0.7 ’9 Rm + (1 + [3)125 3.33 +{76)(033) = 0.0387 m =5 15 = 2.90 m For 15‘ = 150 . I = 4.24-6—03 "3 333+(151)(o.33) Then IC =3.10mA Spacifications arc met. = 0.020? M Rm = RlflR, = 3&1: = 2.4 m R1 _ 12 _ Vm=[R1+R1)Vcc_[12+3](20)—16V (a) For {3:75 20 =(1+ muck. + vflogn) + I” R," + v”, 20— 0.7 — 15 = I,Q[(?6)(2) + 2.4] 50 1,9 =0.0214 mA, {CE = [.60 mA. rm =15: mA Vm =20-(l.6)(1)—(1.62)(2) 01' vm = 15.16 v (b)Forfi=100,wcfind I,Q=0.OI6|nm. Icazmlm. Vs =IS.I3V Electronic Circuit Analysis and DesignI 2"" edition Solutions Manual 3.44 Icy;- =4.B mA - 15;; = 4.84 mA Vase = Vac - Icoflc - 15133.5 5 =13 — (4.311(2) — (4.34m; => RE = 0.496 H1 RT" = (0.1“). + ,6)R£ =(0JHI21MOAQG) = 6.0 kn Vrg = IBQRTH + VBE(°“) + [1 + ENBORE‘ l I V = .._. R .V = _ _ TH R1 TH cc- Rlfli CHIS) 31—10mm =(o.04){s.0) + 0.70 + (121){o.04)(0.496) 1 if{(1011) = 3.34 32.31%; = . kfl. —--—- = . 33—33—- 320 + R: 6 0 R: = 7.37 kfl 3.45 For nominal .3 = T "J ng = fi = 0.0235 mA - [so = 2.03 m."|. Vcsa = Vc'c: - Ichc -- 15133.5: 10 = 20 — (2)14) — {2.03335 9 RE = 0.935 K RTE = (0.1)“. + flute = (0.1}(71}{D.955} = 6.99 K V73 = ngRrH + ng(on) + ngfig 1 .. _ R—I - Rn: - 'ch = 5205171; + V5503“) + 15035 Ril(5.99)(20) =(0.0236)(6.991 + 0.70 + (2.033(0935) 1 R71139.11} _. 2.90 $8.21!: R = _2 ‘ _._.._ = _ -‘————“ K 43.2 + R, 6 99 R3 = 3.18 K Chuck: For .3 = 50 1.13 v _—, _.__ = m (“a +“Hg-Jan) 2.90 Vm - Vagina) _ 2.90 — 0.7 I = —-——————— _ _._.._..___ 9" Rn: + (1 +533; 5.99 + (511(0935) = 0.0334 mm Icq = 1.92 mA For )5 = 90 La - fl — o 0223 A ° ‘ c.99+(91)(0.935) ‘ ' m 1:2 = 2.05 mA D. ..nu.al 3.45 [cg =1 mA at ng =1.02 mA V's-sq = Vcc - Ichc - IEQRE. 5 =15 — (1)15} — (1.02135 => RE = 4.90 kn Bias stable: - RT” = (0.1)(1 +3)R5 == (0.1)(61](4.9] = 29.9 kn 1 {HQ = 63 = 0.0167 mA 1 . VTH .._. ETH- - Vac = IBQRTH + I"BE{cm') +IEC|RE Tel-(29.91115) 40015702901 + 0.70 1 + (1.02)(4.901 Laws) = 5.197 R1 72.43: _ R] - 72.4 kn, 711+ R7 — 29.9 R: = 50.9 kn CheckrFo:5:-.45 50.9 v = ——--- ’ =. v 7" (50-9+72.4)”°) “9 WI'H -V3£(on) 6.19 —0.T 13° = m = m = 0.0215 0131 Icq = 0.953 319.. 91-? = 3.27. Chad: For .3 = 75 Ian = mm— = 0.0136 m2. 29.9 + (75)(4.90) Ice =10: mA. —° -.-.- 2.0% It: Design crilcrion is satisfied. 3.47 (a) Vcc 5 1.31025 + R£)+vm 3 = (0.1)(512'E +RE)+1_4 :9 R5 = 2.57 11': RC = 13.319 , 1,0 =%=0333m R,” = (0.1)(1+ ma. =(0.l)(121)(2.67) = 32.319 1 1 Vm = F'RTH "vac =F(32'3)(3) I I = “QR”, +V,£(on)+(1+ fl)I,QRE = (0.000833)(31.3)+ 0.7 + (121X0.00083 3X16?) which gives Rl = 973 El, and R2 =4S.4 m ___—_..__ Chagter 3: Problem Solutions 3.48 3 .49 3.50 3 3 (b) 1',[ E R1 HQ: zm=>20fim rm=1mm P = (im +1.11% =(l00+?.0.6)(3) 01' P=362 ,uW 1525—;‘E-§=1.67m R5 - 3 RT}! = R1IIR: = (0.1}(1 + )3}RE = (0.1](101)(3) = 30.3 k0 R: 1 v73-(R1+R2)(4)—2-E'RTH-(4)—2 Isa 1+5 I50 = = 0.0165 nut 5 = ngR3 + V2.9(on) + [BETH + V”: 5 = (1.5T}{3] + 0.7 + [0.0165)(30.3) 1 + R—‘(30.3}[1) - 2 0.30 = $430.3)“; => 5:, :15: m 1 152R; _— = {13 = _ 152+R: 3 = R3 3781:!) 5- Rm = Rle-R: =10[|20 =5 Rm = 6.67 m . _ R: 20 1'” " (It. +Rg)(1°) ‘5 = (20 +1::h)(ml ‘5 =:v VIE = 1.57 V b. 10=II+BJIBQIR£+VE£(OII)+IEQRrH-FVTH I3 = 10—03—157 _ 7.53 ° 5.s7+[51)(2) _ 123.? => Iaq = 0.0593 LuA Ice = 3.65 EDA, Isa = 3.52 mA V5 = to - 15°35 =10 — {3.52)(2) V; = 2.76 V V: = ICQRC — 10 = (3.56)(2.2) — 10 'c = -2.17 v V‘ -V‘ :-: [camc -t-RE)+V5m 20: (OSXRc + R£)+8=:(RC+R£)= 24 in Let RE=10kfl than Rc=l4kfl be! ,8 = 60 from provious problcm. R1." = (0.1)(l “Sm. = (0.1)(61}(10) 351 0r Rm=61kfl 1,0 =% ='O.DOS33mA R 1 V = 2 l0 -5=-—-R -10-5 m [ R1 + R2} J R‘ m Now IO = (1+fi)!80RE +V55(0")+ Irmg‘zem +VTH 10 = (61)[0.00833)(10) + 0.7 + (0.00333)(61) l +— 61 10 —5 RI( X ) Then R,=70.0k.Q and 31:47419 - 10 _‘ l0 Rl + R2 70 + 474 So the 40 1% current limit is met. In Ill :alSAM L a”. = mug, = 351120 => 13:12.7 kfl R: _ _ 20 7_- VrH=(RI+R2)(‘)—5-(20+35)(} 3 => VT}; = —2.45 V 13. V71; - Wash“) -[-10) RT}; + (1 + MR: —2.45 - 0.7 + 10 " 12.7 + (76)(D.5] =- Im; = 0.135 $139.. Lea = 10.3- In). I59: ICE 2 10.1 m. chq = 20 — J’ch; — {was = in — (10.1)[0.3) — {10.3)(u.5) vcm = 5.77 v C. R: =20+5%=21kfl R1 = 35 - 5% = 33.25 kn R3 = 0.5 — 5% = OATS k0 R1”: = RfilR: = 2133.25 =12.9 kn 3: V-rH— (R1+R:)[7} 5 2] 2 (21+ 33.25)ml " 5 " ‘2'" V —2.29 - 0.7 - (—10) W Icq =10.T mA. [sq =10.9 111A 15¢: =D.143mA For R:- = 0.3 + 5% = 0.84 kn cho = 20 - [1o.7)(u.34} - {10.9}(o.475) : chq = 5.5-3 V Electronic Circuit Analysis and Design. '2"d edition Solutions Manual For Rc = 0.8 - 5% = 0.75 kn Vase = 20 - (10.7}(035) — {10.9}(o.475) =3- Vcsq = 6.69 V R1=30-5%=l9kn R: = 35 + 55’: = 36.75 kn R9 = 0.5 + 5% = 0.525 kn R”; = {7.1”}!2 =19|i36.75 =12.s m 19 VI. _— _._..._.._ 7 _ - _ _2‘ v H (19+35.75)() 3' 61 —2.t‘:l —n.7 - (an) 12.5 + (757(05sz [ca 2 m5. [£0 = mA {sq m = 0.128 mA For R: = 0.84 kn VCEQ = 20 - (9.53)(0.84) - (9.703(0525} 3 chq = 6.36 V For H: = 0.76 kn Vc'gq = 20 - (9.55)(o.76) — (9.70)[0.525) : Vcsq = 7.63 V 50 9.53 g Icn S 10.7 mm and 5.83 5 chq 5 7.63 V 3.52 L an, = sun musoo mum m = 256 mum m =7? REE = 5%.? 160 .5“er 3-VTH_VTg—!-S) 500 500 ' 7o 5 3 5 1 1 1 500 + 500 - 7n -"”(§7+afi+ a) -- 0.0554 = VTH(O70153} VIE = —3.03 V b. [so = VTH -V55(an) -- (-5) R73 = (1 + #3135 _ 4.03 — 0.7 + s ' 54.7 +{101)[5} Isa = 0.00227 mA Ice = 0.227 mA. 15¢ = 0.229 Page = 20 - (0.227)(50) - (0.2mm cho = 7.51 v 3.53 3.54 Rn; = JDIIGDQIQU =- Rzfl =10 kn 5-VTH 5-V7'H =VTH 30 60 2D 5+5 _V 1+1+1 30 so ‘ T” 30 so 20 ’1 For 3 = 100 V7}: - Vasfon] — {—5) RTE + {1 + {31195 _ 2.5 0.? 4-5 ‘ 10 + (101M031 [sq = FHA. [Cg = 32.5 mA. [.59 = 32.7 m}. Vcsq =15 — (22.5]10.5) — 22.7mm) In sanitation ng= VCEQ = ~0.79! =3- VCEQ = 0.2 V Vs = VT}! - IBQRTH - VEE{°'1) = 2.5 -- (0.225H10) - 0.7 ’5 = —U.45 V :- Vc = -U.I15+ 0.2 = —0.'L'5 V 10 — (—0.25; I” = 0.5 =>' Icq = 20.5 mm Rm = R: HR: = 100nm = 23.5 m R2 40 v _ — = _ .. _ TH (R;+R:)(m] (4fi+1no)(m) 286V Vm - V32{on) 2.35 - 0.7 Rm +(1 Hanan = 23.5 + (1mm IE. = 0.0144 mA I37. = IQ, 21.73 wt, IE; =1.75 mA 10 — v —3—52 = In + In: I _ V5; — VEEODDJ — (—10) E1 — quh—‘F' lfl—Vag _I V32—0.7+10 3 ' “ (121)(5) m 9.3 1 1 — — .7 - — = v - 3 1 3 605 3’ (a + (121m)) 1.59 = Vm(o.3as) =3» V3, = 4.75 v 1.75 — 0.7 — (“10] 152= 5 =Ifl=2.31mA IE; = 0.0232 mA 1;; = 2.79'mA Vase: = 4.75 - {1.75)(1) =3 V5301 = 3.0 v chq2 = 10 — (-4.75 - 0.7) = Vega: = 5.95 V —-————-—————-——-_.___._. Chapter 3: Problem Solutions 3-55 3.5? RT}; = R1 "R: = SUIIIDD = 33.3 kn .. _._ R: lem=-o.T—.07=—14 “1+3” _].4_[_5 ‘ m 100 ) _ _ 7v I52="_1_:I§1=3.5 mA _(100+50 (In) 5—1'5 gm = 0.04.“ mA 5 = [51321 + VEB(°n) '4’ I51 RTE + VTH In 2 3.50 mA 10] = — . = 0.503 IE. (my) a) m 151: 1}“ + 151 = 13! = [ILA 151 = 0.259 mA 5 = (9.505333! + 0.7 + [0.008}{33.3) + 1.67 IE; = 0.00320 mA R5, = 2 93 IL“ [:1 = 0.256 mfi. vi, = s - (0.003)(2.53} = 2.63 v 3.56 V51 = V31 - Vgcq: = 2.63 - 3.5 =2 -0.37 V Current through 1" source 2 15: + £2: and V“ = “0‘87 _ 0'70 = 4‘57 v _ "7... — I51 = 12:2 = (1+ BUB: = (51)(B.26} pA 13; = iris-i]— =0.803 :9 RE; = 4.25 kfl So total cum-.01: 2(51)(B.26) 4023. = 843 uA P' = I: IV'| = (0.3mm => ' = 4.22 mw Vcam = 4 => ch = -1-57 + 4 = 2-43 V (me V" source) RC: = 5 31'“ =:~ RE, = 3.21 m From Example 3.15.14, = 0413 um IRm = Ic, - 15.: = 0.0 — 0.003 = 0.792 111A _ 50 _ —0.07—(—5} _ 50 Ice —- (51)(0.413) = 0.405 mA Rm - —-——-D_m =- M P" = I - W = (0.405}(5) => 2+ = 2 03 mW (From V+ sauna) ...
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chap003 - .EICCtronic Circuit Analysis and Design....

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